If two particles A and B of charges 1.6 times | vedclass

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(C) According to Coulomb's law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k \f

Vedclass · Accepted Answer

$5.12 	imes 10^{-25} \ N$

Vedclass · Answer

(C) According to Coulomb's law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k \frac{|q_1 q_2|}{r^2}$,where $k = 9 	imes 10^9 \ N \cdot m^2/C^2$.Given:$q_1 = 1.6 	imes 10^{-19} \ C$$q_2 = 3.2 	imes 10^{-19} \ C$$r = 3 \ cm = 3 	imes 10^{-2} \ m$Substituting the values:$F = \frac{(9 	imes 10^9) 	imes (1.6 	imes 10^{-19}) 	imes (3.2 	imes 10^{-19})}{(3 	imes 10^{-2})^2}$$F = \frac{9 	imes 10^9 	imes 5.12 	imes 10^{-38}}{9 	imes 10^{-4}}$$F = 5.12 	imes 10^{9-38+4} \ N$$F = 5.12 	imes 10^{-25} \ N$.

If two particles $A$ and $B$ of charges $1.6 \times 10^{-19} \ C$ and $3.2 \times 10^{-19} \ C$ respectively are separated by a distance of $3 \ cm$ in air,then the magnitude of electrostatic force on particle $A$ due to particle $B$ is

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